A coulomb is a measure of charge and current is the rate of flow of charge.
There is a formula linking the three quantities (charge=Q; T=time; I=current):
Q=I x T -> 1.5A x 0.1s= 0.15C
pass
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AC can pass through a capacitor. The higher the frequency of AC the lower the reactance (like resistance). The current and applied voltage are 90 degrees out of phase the current leading the voltage by this amount.
Yes and no. A capacitor generally does not pass DC current, except for a small "leakage current", but upon the inital application of a DC voltage, the capacitor will pass current until it reaches the full potential of the applied voltage. The simple answer is no it does not. In fact we use that characteristic to "decouple" one circuit from another in amplifiers for example.
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Here we are given 3.1 amperes of current and are asked to find the time it takes 10 coulombs of charge to pass a given point. First ask yourself how many coulombs are passing a given point in one second. If we have 3.1 amperes of current, we have 3.1 coulombs of charge passing any given point in one second. If it takes 1 second for 3.1 coulombs of charge to pass, how long will it take for 10C of charge to pass?
If you divide the charge by the time, you get the average current (in amperes).
Current = charge/time = 10/5 = 2 amperes
(4 coulombs / 2 seconds) = 2 coulombs per second = 2 Amperes.
2 amperes
Electrical current is the number of elementary charge units (coulombs) that pass by a given point in one second. Current, measured in amperes, is coulombs per second. Electrical voltage is the "pressure" behind that current. Voltage, measured in volts, is joules per coulomb.
Divide the coulombs by the amperes. The answer will be in seconds. The resistance is irrelevant in this problem.
72 coulombs in 24 seconds is 3 amperes.One ampere is one coulomb per second.
1.22x10^4c
300A means that 300 coulombs of electrons are passing through it per second. 300 coulombs is (1.87 * 10^19) electrons, or simple 300C of electrons.
Current = charge / time Charge q = n * e e = 1.602 x 10^-19 C time given = 0.2 s Current = 0.5 A So I = n e / t Hence n = I * t / e Plug I, t and e. You would get required n ANS: 6.24 x 10^17 electrons
That is 6 amps. One coulomb passing a point once a second is defined as one amp. Take the number of coulombs that have passed the point and divide it by the time it took the charge to pass (in seconds) and you have the current in amps.